Q9:

Prove that $\cot^{-1} (\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}) = \frac{x}{2}$, $x \in (0, \frac{\pi}{4})$

Step 1: Simplifying the Radicals

To simplify expressions involving $\sqrt{1 \pm \sin x}$, we utilize the fundamental trigonometric identities: $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin x = 2\sin(x/2)\cos(x/2)$. [Per the Pythagorean Identity and Double Angle Formula].

We can rewrite the term inside the radical as:

$1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$

$1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2$

Step 2: Applying Square Roots

Given $x \in (0, \pi/4)$, we know that $x/2 \in (0, \pi/8)$. In this interval, $\cos(x/2) > \sin(x/2) > 0$.

Therefore:

$\sqrt{1 + \sin x} = \sqrt{(\cos(x/2) + \sin(x/2))^2} = \cos(x/2) + \sin(x/2)$

$\sqrt{1 - \sin x} = \sqrt{(\cos(x/2) - \sin(x/2))^2} = \cos(x/2) - \sin(x/2)$

Step 3: Substitution into the Expression

Let the argument of the cotangent inverse be $A$. Substituting the results from Step 2:

$A = \frac{(\cos(x/2) + \sin(x/2)) + (\cos(x/2) - \sin(x/2))}{(\cos(x/2) + \sin(x/2)) - (\cos(x/2) - \sin(x/2))}$

Simplifying the numerator and denominator:

$A = \frac{2\cos(x/2)}{2\sin(x/2)} = \frac{\cos(x/2)}{\sin(x/2)} = \cot(x/2)$

Step 4: Final Inverse Operation

The expression now becomes:

$\cot^{-1} (\cot(x/2))$

Since $x/2 \in (0, \pi/8)$, and the principal value branch of $\cot^{-1}$ is $(0, \pi)$, we apply the property $\cot^{-1}(\cot \theta) = \theta$ for $\theta \in (0, \pi)$. [Per the definition of Inverse Trigonometric Functions].

Therefore, $\cot^{-1}(\cot(x/2)) = x/2$.

Geometric Representation of Identity

Final Solution: $\cot^{-1} \left( \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \right) = \frac{x}{2}$